Question

If ${\left({\tan }^{-1}x\right)}^2+{\left({\cot }^{-1}x\right)}^2=\frac{5{\pi }^2}{8}$, then $x$ equal to

• A. $-1$
• B. $1$
• C. $0$
• D. $2$

Answer 1

The correct option is A $-1$

Consider the given equation.

${\left({\tan }^{-1}x\right)}^2+{\left({\cot }^{-1}x\right)}^2=\frac{5{\pi }^2}{8}$

Since, ${\cot }^{-1}x=\frac{\pi }{2}-{\tan }^{-1}x$

Therefore,

${\left({\tan }^{-1}x\right)}^2+{(\frac{\pi }{2}-{\tan }^{-1}x)}^2=\frac{5{\pi }^2}{8}$ ……. (1)

Let $t={\tan }^{-1}x$

From equation (1),

$t^2+{(\frac{\pi }{2}-t)}^2=\frac{5{\pi }^2}{8}$

$t^2+\frac{{\pi }^2}{4}+t^2-\pi t=\frac{5{\pi }^2}{8}$

$2t^2-\pi t=\frac{5{\pi }^2}{8}-\frac{{\pi }^2}{4}$

$2t^2-\pi t=\frac{3{\pi }^2}{8}$

$t^2-\frac{\pi }{2}t-\frac{3{\pi }^2}{16}=0$

$t=\frac{\frac{\pi }{2}\pm \sqrt{\frac{{\pi }^2}{4}-4\times 1\times -\frac{3{\pi }^2}{16}}}{2}$

$t=\frac{\frac{\pi }{2}\pm \sqrt{\frac{{\pi }^2}{4}+\frac{3{\pi }^2}{4}}}{2}$

$t=\frac{\frac{\pi }{2}\pm \sqrt{\frac{4{\pi }^2}{4}}}{2}$

$t=\frac{\frac{\pi }{2}\pm \pi }{2}$

$t=-\frac{\pi }{4},\frac{3\pi }{4}$

Therefore,

${\tan }^{-1}x=t$

$x=\tan (-\frac{\pi }{4})\Rightarrow -1$

$x=\tan \left(\frac{3\pi }{4}\right)\Rightarrow -1$

Hence, the value of $x$ is $-1$.